Let $G=\operatorname{Gal}(L/K)$ and let $\mathcal P,\mathcal Q$ lie over $\mathfrak p$. These primes define two extensions to $L$ of the $\mathfrak p$-adic absolute value on $K$. The <conjugacy of extensions of a valuation to a normal extension> says that some $\sigma\in G$ carries the first extension to the second. Equivalently, $\sigma\mathcal P=\mathcal Q$. This proves the <transitivity of the Galois action on primes>.

The <decomposition group> is the stabilizer
$$
G_{\mathcal P/\mathfrak p}
=\{\sigma\in G:\sigma\mathcal P=\mathcal P\}.
$$
Each such automorphism extends continuously to the completions, and restriction gives the <decomposition group of a prime and local Galois group> isomorphism
$$
G_{\mathcal P/\mathfrak p}\cong
\operatorname{Gal}(L_{\mathcal P}/K_{\mathfrak p}).
$$

For the splitting field $L=\mathbb Q(\sqrt[3]7,\zeta_3)$ of $X^3-7$, the global Galois group is $S_3$. Since $7\equiv1\pmod3$, the field $\mathbb Q_7$ contains $\zeta_3$. The local splitting field is therefore $\mathbb Q_7(\sqrt[3]7)$, an Eisenstein, totally ramified cyclic extension of degree three. Hence there are
$$
[S_3:C_3]=2
$$
primes of $L$ above $7$, and the decomposition group of each is the normal subgroup
$$
A_3\cong C_3.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 136 / 4 / b / Solution

{title2=$G_{\mathcal P/\mathfrak p}$}
{wiki}

For a prime $\mathcal P$ of a finite Galois extension $L/K$ above $\mathfrak p$, the decomposition group is the stabilizer
$$
G_{\mathcal P/\mathfrak p}
=\{\sigma\in\operatorname{Gal}(L/K):\sigma\mathcal P=\mathcal P\}.
$$


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